Messages in this thread | | | Date | Tue, 16 May 2017 15:19:03 -0700 | From | Davidlohr Bueso <> | Subject | Re: [PATCH 2/6] locking: Introduce range reader/writer lock |
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On Mon, 15 May 2017, Peter Zijlstra wrote:
>On Mon, May 15, 2017 at 02:07:21AM -0700, Davidlohr Bueso wrote: > >> + * Fairness and freedom of starvation are guaranteed by the lack of lock >> + * stealing, thus range locks depend directly on interval tree semantics. >> + * This is particularly for iterations, where the key for the rbtree is >> + * given by the interval's low endpoint, > > >So suppose the lock is held at [a,n], and I want to acquire [g,z], this >conflicts, therefore I wait.
Ok, then, ref [g,z] = 1, ref [a,n] = 0 (lock owner). Per below, at this point the tree will overlap with anything between [a,z], which is the world.
> >While I wait, someone else comes in at [b,m], they too wait.
[b,m] intersects with both nodes above, thus ref [b,m] = 2.
> >[a,n] is released, per ordering [b,m] acquires, I still wait.
Now: ref [g,z] = 0 ref [b,m] = 1
So due to reference counting [g,z] is acquired, despite [b,m] being _put() before [g,z].
>[a,n] returns to wait.
Similar to the [b,m] case, when [a,n] comes back, it too will get a ref = 2 and hence "go back in line". Iow, lock order does depend on have fifo semantics among contended ranges.
> >[b,m] releases, does the iteration then restart and grant it to [a,n] or >will I (at [g,z]) finally acquire? > > >Since the code always does range_interval_tree_foreach() it would appear >to me [b,m] will always win and [g,z] could be made to wait >indefinitely (by always contending with another range that has a lower >starting point). > > > >> and duplicates are walked as it >> + * would an inorder traversal of the tree. > >Are duplicates ordered in FIFO ? Afaict the above is free of actual >semantics.
This will strictly depend on the rotation when you have duplicates when more nodes are added later. But again that's the order of walking the tree.
Thanks, Davidlohr
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